Answer
$6\sqrt{3}$
Work Step by Step
$y=\sec^2 x$, $y=8\cos x$
Find the intersections:
$\sec^2 x=8\cos x$
$\cos^2 x\sec^2 x=\cos^2 x*(8\cos x)$
$1=8\cos^3 x$
$\cos^3 x=\frac{1}{8}$
$\cos x=\frac{1}{2}$
$\cos x=\frac{\pi}{3}+2k\pi, -\frac{\pi}{3}+2k\pi$
It is given in the problem that we only want the area between $x=-\frac{\pi}{3}$ and $x=\frac{\pi}{3}$. Note that between $-\frac{\pi}{3}$ and $\frac{\pi}{3}$ $8\cos x>\sec^2 x$, so the integrand is $8\cos x-\sec^2 x$.
The area is:
$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\cos x-\sec^2 x)dx$
$=(8\sin x-\tan x)\left|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\right.$
$=(8\sin\frac{\pi}{3}-\tan\frac{\pi}{3})-(8\sin(-\frac{\pi}{3})-\tan(-\frac{\pi}{3}))$
$=(8*\frac{\sqrt{3}}{2}-\sqrt{3})-(-8*\frac{\sqrt{3}}{2}-(-\sqrt{3}))$
$=4\sqrt{3}-\sqrt{3}+4\sqrt{3}-\sqrt{3}$
$=6\sqrt{3}$
